"Cristian Baboi" <[EMAIL PROTECTED]> wrote: > On Fri, 11 Jan 2008 09:11:52 +0200, Lennart Augustsson > <[EMAIL PROTECTED]> wrote: > > > Some people seem to think that == is an equality predicate. > > This is a big source of confusion for them; until they realize that > > == is just another function returning Bool they will make claims > > like [1..]==[1..] > > having an unnatural result. > > The == function is only vaguely related to the equality predicate > > in that it > > is meant to be a computable approximation of semantic equality (but > > since it's overloaded it can be anything, of course). > > > > >> So let's imagine: > >> > >> ones = 1 : ones > >> > >> ones' = repeat 1 > >> where repeat n = n : repeat n > > (==) :: Eq a => a -> a -> Bool > > -- what is (y (y) ) by the way ? > -- how about ( y id ) ? > > y f = f (y f). > > ones :: Num a => [a] > ones = y (1 :) > > repeat :: a -> [a] > repeat = \n -> y (n:) > > ones' :: Num a => [a] > ones' = repeat 1 = (\n->y(n:)) 1 = y (1 : ) > > To be able to test them for equality, we must have Eq a. > > So, the reason we cannot test them for equality is that we cannot > test y (a : ) == y (a : ) where a == a is testable. > Yes, thanks. I actually do think that many things would be easier if every recursion would be translated to its fixpoint, making the term tree completely finite and defining y internal, as it's arcane, black magic.
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